Evaluate the iterated integral. $ \int_0^\pi \left( \int_2^{4} xy\cos(y) \, dx \right) dy =$ Choose 1 answer: Choose 1 answer: (Choice A) A $4 \pi$ (Choice B) B $-12$ (Choice C) C $-2$ (Choice D) D $0$
Evaluate the inner integral: $\begin{aligned} &\int_0^\pi \left( \int_2^{4} xy\cos(y) \, dx \right) dy \\ \\ &= \int_0^\pi \left[ \dfrac{x^2y\cos(y)}{2} \right]_2^4 dy \\ \\ &= \int_0^\pi y\cos(y) \left( \dfrac{16}{2} - \dfrac{4}{2} \right) dy \\ \\ &= 6 \int_0^\pi y\cos(y) \, dy \end{aligned}$ We'll use integration by parts to evaluate the outer integral. Let $u = y$ and $dv = \cos(y) \, dy$. Then $du = dy$ and $v = \sin(y)$. $\begin{aligned} & 6 \int_0^\pi y\cos(y) \, dy \\ \\ &= 6y \sin(y) \bigg|_0^\pi -6 \int_0^\pi \sin(y) \, dy \\ \\ &= 6(\pi \sin(\pi) - 0 \sin(0)) + 6\cos(y) \bigg|_0^\pi \\ \\ &= 0 + 6(-1 - 1) \\ \\ &= -12 \end{aligned}$ The answer: $ \int_0^\pi \left( \int_2^{4} xy\cos(y) \, dx \right) dy = -12$